An ideal gas is taken from the state A (Pressure P, volume V) to the state B (Pressure P/2, volume 2V) along a straight line path in the P-V diagram. Select the incorrect  statement(s) from the following:

(i) Work done = area under the curve
    ${\mathrm{W}}_1=\left(\mathrm{P}+\frac{\mathrm{P}}{2}\right)(2\mathrm{V}−\mathrm{V})=\frac{3}{2}\mathrm{PV}$
We know that work done under isothermal process
   $\begin{array}{l}{\mathrm{W}}_2=\mathrm{RT}×2⋠3026\log ⁡(2\mathrm{V}/\mathrm{V})\\ {\mathrm{W}}_2=0⋠693\mathrm{RT}=0⋠693\mathrm{PV}\\ {\mathrm{W}}_1>{\mathrm{W}}_2\left(1\right)\text{ is correct. }\end{array}$
(ii) Let P₀ and V_o be the intercepts on the P and V axes.
      Now the equation of straight line would be
   $\mathrm{P}=−\frac{{\mathrm{P}}_0}{{\mathrm{V}}_0}×\mathrm{V}+{\mathrm{P}}_0(âˆ\mu \mathrm{y}=\mathrm{mx}+\mathrm{C})$
Further, $\mathrm{PV}=\mathrm{RT}\text{ or }\mathrm{P}=(\mathrm{RT}/\mathrm{V})$
            $\frac{\mathrm{RT}}{\mathrm{V}}=−\frac{{\mathrm{P}}_0}{{\mathrm{V}}_0}×\mathrm{V}+{\mathrm{P}}_0$
This represents a parabola. hence (2) is correct.

$$\begin{gathered} \left(iii\right) PV =RT (for one mole) also P = P_0-\frac{P_0}{V_0}×V \\ Combining the above equations , RT=P×\left[\frac{V_0}{P_0}\right(P_0-P\left)\right] , which represents a parabola. \\ So option \left(3\right) is incorrect . \end{gathered}$$
(iv) Here P - V  graph deviates most from the isotherm at the mid point ($\frac{3P}{4},\frac{3v}{4})$
So, the temperature of the gas is maximum at the state point ($\frac{3P}{4},\frac{3V}{4}).$. Hence (4) is correct.